Suppose i have a bucket full of water in a closed room,the water is

suppose i have a bucket full of water in a closed room,the water is kept undisturbed for say 5 days,after those 5 days i found that the temperature of water was lower than as it was 5 days before and so it does for the room
all this (for the system) isn't going in accordance with second law of thermodynamics
where i am wrong(if i am)?

I believe the temperature of the room will increase. As you mentioned it a closed room and the heat lost by the bucket of water has to go in increasing in temperature of the room. This would be in accordance with the second law as now the energy is distributed among the enormously large (compared to that of water in the bucket) degrees of freedom of the gas molecules of the room.

Staff: Mentor

suppose i have a bucket full of water in a closed room,the water is kept undisturbed for say 5 days,after those 5 days i found that the temperature of water was lower than as it was 5 days before and so it does for the room
all this (for the system) isn't going in accordance with second law of thermodynamics
where i am wrong(if i am)?

If the room is truly isolated thermodynamically, and if the bucket and the dry air were initially at the same temperature then you would indeed find that the bucket of water and the air were at a lower temperature 5 days later.

So, since we know that the SLOT is correct, how did entropy increase in a situation where the temperature decreased and no heat was exchanged with the surroundings?

If the room is truly isolated thermodynamically, and if the bucket and the dry air were initially at the same temperature then you would indeed find that the bucket of water and the air were at a lower temperature 5 days later.

So, since we know that the SLOT is correct, how did entropy increase in a situation where the temperature decreased and no heat was exchanged with the surroundings?

Staff: Mentor

Re: entropy

Yes, I know. Think about it a bit and try to figure out the answer. Make a guess under the assumption that physicists have encountered buckets of water and designed the SLOT accordingly. You might surprise yourself and find out that you know physics better than you suspect.

Try first applying FLOT - the first law of thermodynamics. That will allow you to determine Q: the heat flow from/to the room and the heat flow from/to the water

How is the change in entropy related to this heat flow and temperature? Once you figure out the heat flows you can work out the entropy change and perhaps then you will be able to see how FLOT fits into SLOT.

obviously the molecules in air,but the point is the downfall in temperature of room?

What direction is the heat flow relative to the water when it vaporizes? What does that tell you about the sign of the change in entropy of the water? What is the direction of heat flow relative to the air in the room? What does that tell you about the sign of the change in entropy of the air? What is the temperature of the water and room while this heat flow is occurring? What does that tell you about the relative magnitudes of the two entropy changes?

The first thing to consider is the KE of a water molecule in the liquid phase at room temperature and the KE of a water molecule in the vapor phase at room temperature. Clearly, even though they are at the same temperature, the KE of the vapor is much higher than the KE of the liquid. This energy is called the heat of vaporization. Because energy is going from the liquid into the vapor phase, the temperature of the liquid phase must decrease.

As long as the air is dry there will be more water molecules going from the liquid to the vapor phase than vice versa. As long as that happens, the temperature of the liquid must decrease in order to supply the heat of vaporization.

OK, so you now understand how the second law of thermo and entropy work in this situation, correct?

The first thing to consider is the KE of a water molecule in the liquid phase at room temperature and the KE of a water molecule in the vapor phase at room temperature. Clearly, even though they are at the same temperature, the KE of the vapor is much higher than the KE of the liquid. This energy is called the heat of vaporization. Because energy is going from the liquid into the vapor phase, the temperature of the liquid phase must decrease.

As long as the air is dry there will be more water molecules going from the liquid to the vapor phase than vice versa. As long as that happens, the temperature of the liquid must decrease in order to supply the heat of vaporization.

okay,suppose at moment say 't' the temperature of room is "T",this is the state of initial equilibrium,the water molecules can vaporize at this temperature,they do so by increasing the K.E of the dry air,this will ,however, can still allow the flow of heat from air to water which cannot happen because the temperature available cannot condense it
it's like the system is disturbed from the initial equilibrium to throw away heat into air?

What direction is the heat flow relative to the water when it vaporizes? What does that tell you about the sign of the change in entropy of the water? What is the direction of heat flow relative to the air in the room? What does that tell you about the sign of the change in entropy of the air? What is the temperature of the water and room while this heat flow is occurring? What does that tell you about the relative magnitudes of the two entropy changes?

Staff: Mentor

Re: entropy

The water molecules "colliding" with the surface of the water can lose their energy to the liquid and become liquid again themselves. In equilibrium the amount of water evaporating and condensing back into the water is equal over time. It is only when you factor in the average amount of molecules evaporating vs the ones condensing that you see this. A snapshot could show that either the air or the water has more or less energy that it does a second later.